Final answer:
To calculate the passenger's weight while the elevator is accelerating, we find the elevator's acceleration and the additional force due to this acceleration. Adding this force to the passenger's normal weight due to gravity gives us the total apparent weight. However, the calculated weight of 799.36 N is not an option in the problem, indicating a possible error in the provided choices.
Step-by-step explanation:
The question asks for the passenger's weight while the elevator is speeding up. To find this, we need to take into account the acceleration of the elevator and the effect it has on the apparent weight of the passenger.
First, we calculate the acceleration 'a' using the formula:
a = (v - u) / t
where:
- v is the final velocity (11 m/s)
- u is the initial velocity (0 m/s, since it starts from rest)
- t is the time taken to reach final velocity (4.1 s)
Thus, a = (11 m/s) / (4.1 s) = 2.68 m/s²
Now, we use Newton's second law (F = ma) to find the additional force (Fadditional) due to acceleration:
Fadditional = m * a
For a passenger of mass 64 kg:
Fadditional = 64 kg * 2.68 m/s²
Fadditional = 171.52 N
The weight of the passenger is the force due to gravity (Fgravity = m * g), where g is the acceleration due to gravity (9.81 m/s²):
Fgravity = 64 kg * 9.81 m/s²
Fgravity = 627.84 N
The total apparent weight of the passenger while the elevator is accelerating is the sum of these two forces:
Total weight = Fgravity + Fadditional
Total weight = 627.84 N + 171.52 N
Total weight = 799.36 N
The closest answer to the passenger's weight while the elevator is speeding up is therefore 800 N, which is not listed in the available options. Since the question does not provide this option, there seems to be an error in the provided choices or the premises of the problem.